Question Bank - Mathematics

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What is the equation of the circle which touches both the axes in the first quadrant and the line y - 2 = 0?

A.
x2 + y2 - 2x - 2y - 1 = 0
B.
x2 + y2 + 2x + 2y + 1 = 0
C.
x2 + y2 - 2x - 2y + 1 = 0
D.
x2 + y2 - 4x - 4y + 4 = 0

Solution:

Concept:The general equation for a circle is (x - h)2 + (y - k)2 = r2, where ( h, k ) is the center and r is the radius.Calculation:The circle which touches both the axes in the first quadrant and the line y - 2 = 0 is shown in the figure.From the above figure, we can say that the diameter of the circle is 2.? r = 1 and the center of the circle is (1, 1).We know that the equation of the circle is (x - h)2 + (y - k)2 = r2, where (h, k) is the center and r is the radius.? (x - 1)2 + (y - 1)2 = 12? x2 - 2x + 1 + y2 - 2y + 1 = 12? x2 + y2 - 2x - 2y + 2 = 1? x2 + y2 - 2x - 2y + 1 = 0? The equation of the circle is x2 + y2 - 2x - 2y + 1 = 0.

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